Psych. 610
Prof. Moore
Handout #18, p. 1
PARTIAL INTERACTIONS
(No longer covered due to time constraints;
Handout included for your future reference)
PROBLEM
H
M
E
YA j
MSS/AP = 2.30
dfS/AP = 48
n=5
!
H
M
E
Tot
L
3
2
2
2.33
AROUSAL
ML
MH H
4
2
2
5
5
2
3
5
4
4.0
4.0 ! 2.67
QUADRATIC COEFFS
L
ML
MH H
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-3
3
3
-3
Tot
0
0
0
YPk
2.75
3.50
3.50
Subjects perform easy, medium, or
hard problems at one of four levels
of arousal: low, medium low,
medium high, and high. The prediction is that each problem will
produce quadratic data, but the
peaks will not necessarily be at the
same place.
We will test using the partial interactions
ArousalQ x Problem
Start by doing separate SSQ for each problem.
SSQ@HARD = 5(-3+4+2-2) = 1.25
4
SSQ@MEDIUM = 5(-2+5+5-2) = 45
4
SSQ@EASY = 5(-2+3+5-4) = 5
4
The sum of these three SS values indicates how much overall “quadraticness” there is in the data set as
a function of arousal level. But it overestimates the SSAROUSALQ x PROBLEM because it includes the part
that is reflected in the marginal means for arousal. Note that the row quadratic coefficient sums equal
zero but the column coefficient sums have the quadratic pattern. Therefore we need to correct for this
overestimate by calculating and subtracting out the SSAROUSAL QUAD.
SSQ = 15(-2.33+4+4-2.67)2 = 33.75
4
Handout #18, p. 2
p
SSA QUAD x P =
"SS
k =1
!
A QUAD@Pk
- SSA QUAD
= 51.25 - 33.75
= 17.5
To test this, we must know how many df we have in the SSA QUAD x P. There are two ways to think
about this.
Method 1: The formula for the df should reflect the formula for the SS. Therefore
dfA QUAD x P = "dfA QUAD@Pk - dfA QUAD
!
=
px1
=
2
-
1
Method 2: The df for an interaction always equals the product of the individual dfs.
DfA QUAD x P = dfA QUAD x dfP
= 1
x (p-1)
= p-1
= 2
So, we have
SSA QUAD x P = 17.5
on 2 df
MSA QUAD x P = 8.75
FA QUAD x P = 8.75/2.30 = 3.80
on (2, 48) df
p < .05
RELATION BETWEEN PARTIAL INTERACTIONS AND TWO-WAY CONTRASTS
The partial interaction idea says there will be a particular pattern on a factor (e.g., quadratic arousal
factor) that will differ as a function of another factor (e.g., problem).
Two-way contrasts make us specify a particular pattern on both factors. How many such different (i.e.,
orthogonal) patterns could we specify on the problems factor if we wanted to? Answer: 2 (since df for
Problems equals 2).
Handout #18, p. 3
Let’s pick two orthogonal contrasts arbitrarily, say problemLIN and problemQUAD and compute the two
two-way contrasts.
SSA QUAD x PLIN
-1
0
1
-1
1
0
-1
Coefficients
1
1
-1
-1
0
0
1
1
-1
1
0
-1
SSA QUAD x PQUAD
-1
2
-1
-1
1
-2
1
Coefficients
1
1
-1
-1
2
2
-1
-1
-1
1
-2
1
=
5(3-4-2+2-2+3+5-4)2
8
=
5(1)2/8
=
.625
=
5(3-4-2+2-4 +10+10-4+2-3-5+4)2
24
=
5(9)2/24
=
16.875
Now we see that
SSA QUAD x P LIN + SSA QUAD x P QUAD = SSA QUAD x PROBLEM
.625
16.875
=
17.5
More generally, SSA COMP x B is the pooled SS that would be found in the subset of orthogonal 2-way
contrasts that specify ACOMP combined with all the different BCOMP’s.
RELATION BETWEEN PARTIAL INTERACTIONS AND SIMPLE MAIN EFFECTS
Recall that
∑SSA@B = SSA + SSAB
We can rearrange this to
Handout #18, p. 4
SSAB = ∑SSA@B - SSA
Note the similarity to the equation for the partial interaction.
SSA QUAD x B = ∑SSA QUAD@B - SSA QUAD
The logic of the two cases is identical, but our focus differs. In the case of simple main effects we
want to divide our 2-way design into 1-way designs, and so we perform an operation that effectively
pools the SS for a main effect and an interaction.
In the case of partial interaction, we are interested in a component of the interaction. Our operations
inflate this by a component of a main effect and we must subtract that out.